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Jordan's inequality (Theorem)

Jordan's Inequality states that

$\displaystyle \frac{2}{\pi}x\leq\sin(x)\leq x $
for all $ x\in [0,\frac{\pi}{2}]$.



"Jordan's inequality" is owned by Koro. [ full author list (2) | owner history (2) ]
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See Also: comparison of $\sin þeta$ and $þeta$ near $þeta = 0$


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proof of Jordan's Inequality (Proof) by mathcam
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This is version 6 of Jordan's inequality, born on 2001-08-13, modified 2007-06-11.
Object id is 13, canonical name is JordansInequality.
Accessed 6263 times total.

Classification:
AMS MSC: 26D05 (Real functions :: Inequalities :: Inequalities for trigonometric functions and polynomials)
Pending Errata and Addenda
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Proof by jack202 by jack202 on 2002-09-23 12:04:53
sin(x) is a convex function in [0;pi/2], so Jordan's inequality
is simply a special case of the inequality

f'(a)(x-a) <= f(x)-f(a) <= (f(b)-f(a))(x-a)/(b-a)

for a f convex and analitic in [a;b]. This inequality
is easily reconducible to consideration about
secant-tangent (Newton method)

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